The Large Sample Performance of a Maximum Likelihood Method for OFDM Carrier Frequency Offset Estimation

Abstract

This paper is concerned with the performance analysis of a blind method for carrier frequency offset (CFO) estimation in OFDM systems. The method generates a maximum likelihood CFO estimate when message-carrying symbols and channel are assumed nonrandom. It has been observed in simulation that, when a large number of OFDM blocks are in use under noisy scenarios, the performance of the method does not achieve the corresponding Cramer-Rao lower bound (CRB) and is lower bounded by another (larger) CRB when symbols are assumed random. This phenomenon is caused by the inconsistency of nonrandom symbol estimates and is further verified by using first-order perturbation analysis and comparison with the two CRBs.

Appendices

Appendix 1: Proof of CRB Expressions

The noisy data model (10) is similar to (1.1a) of [12] where $\mathbf{E} \mathbf{W} z_{\psi }^{(k-1)(N+N_{g})}$ plays an equivalent role as the array manifold matrix $A$ and $\tilde{\mathbf{s}}(k)$ is equivalent to amplitudes $x(t)$. However, the dependence of $\mathbf{A}$ on $\psi$ is different, with all columns of $\mathbf{A}$ being functions of $\psi$. This difference leads to varied expressions to (E.8f) and (E.8g) in [12] under the conditional model and to the first-order partial derivative of the covariance matrix in (16) of [14] under the unconditional model.

Proof of Lemma 2

For CFO estimation, the diagonal matrix $X(k)$ (for amplitudes) in (E.8f) and (E.8g) should be replaced by the column vector $\tilde{\mathbf{s}}(k)$. Then following the derivation of $CRB^{-1}(\theta )$ after (E.11b), the Fisher information matrix for the CFO becomes a scalar which is equal to

$$\begin{aligned} F_{c}= \frac{2}{\sigma ^{2}} \sum _{k=1}^{K} \tilde{\mathbf{s}}(k)^{H} \mathbf{Q}^{^{\prime }H} \mathbf{P}_\mathbf{Q}^{\perp } \mathbf{Q}^{^{\prime }} \tilde{\mathbf{s}}(k) = \frac{2K}{\sigma ^{2}} tr [ \mathbf{Q}^{^{\prime }H} \mathbf{P}_\mathbf{Q}^{\perp } \mathbf{Q}^{^{\prime }} \hat{\mathbf{R}}_{\tilde{\mathbf{s}}} ] \end{aligned}$$

(49)

where $\mathbf{Q}=\mathbf{A} z_{\psi }^{(k-1)(N+N_{g})}$. Since $\mathbf{R}_{\tilde{\mathbf{s}}}$ is assumed to be positive definite, for $K >>1, \hat{\mathbf{R}}_{\tilde{\mathbf{s}}}$ can also be considered positive definite and hence $tr [ \mathbf{Q}^{^{\prime }H} \mathbf{P}_\mathbf{Q}^{\perp } \mathbf{Q}^{^{\prime }} \hat{\mathbf{R}}_{\tilde{\mathbf{s}}(k)} ]$ is positive according to Lemma 1. Thus one can write

$$\begin{aligned} CRB_{c} = F_{c}^{-1} = \frac{\sigma ^{2}}{2K} \left( tr \left[ \mathbf{Q}^{^{\prime }H} \mathbf{P}_\mathbf{Q}^{\perp } \mathbf{Q}^{^{\prime }} \hat{\mathbf{R}}_{\tilde{\mathbf{s}}(k)}\right]\right)^{-1}. \end{aligned}$$

(50)

Using (9), one can have

$$\begin{aligned} \mathbf{P}_\mathbf{Q}^{\perp }&= \mathbf{I}-\mathbf{A} (\mathbf{A}^{H} \mathbf{A})^{-1} \mathbf{A}^{H} = \mathbf{I}-\mathbf{A} \mathbf{A}^{H}\end{aligned}$$

(51)

$$\begin{aligned} \mathbf{Q}^{^{\prime }}&= \mathbf{A}^{^{\prime }} -j(k-1)(N+N_{g}) \mathbf{A}. \end{aligned}$$

(52)

Using $\mathbf{Q}^{^{\prime }H} \mathbf{P}_\mathbf{Q}^{\perp } \mathbf{Q}^{^{\prime }} = (\mathbf{A}^{^{\prime }})^{H} (\mathbf{I}_{N} - \mathbf{A} \mathbf{A}^{H}) \mathbf{A}^{^{\prime }} = \mathbf{H}$ ³, the conditional CRB can be simplified into

$$\begin{aligned} CRB_{c} = \frac{\sigma ^{2}}{2K} ( tr[ \mathbf{H} \hat{\mathbf{R}}_{\tilde{s}} ] )^{-1}. \end{aligned}$$

(53)

Proof of Lemma 3

Under the unconditional model, the Fisher information matrix (FIM) has the same form as (4) in [14] where the covariance matrix of the data should be replaced by $\mathbf{R}$ as defined in (19). Since all the columns of $\mathbf{A}$ depend on $\psi$, the first-order partial derivative of the covariance matrix assumes the following expression (as opposed to (16) in [14])

$$\begin{aligned} \frac{\partial \mathbf{R}}{\partial \psi } = \mathbf{A}^{^{\prime }} \mathbf{R}_{\tilde{s}} \mathbf{A}^{H} + \mathbf{A} \mathbf{R}_{\tilde{s}} (\mathbf{A}^{^{\prime }})^{H}. \end{aligned}$$

(54)

Define the first-order partial derivative vector of $vec( \mathbf{R})$ with respect to $\psi$ as

$$\begin{aligned} \mathbf{g}=\left[ ((\mathbf{R}^{-1})^{1/2})^{T} \otimes (\mathbf{R}^{-1})^{1/2} \right] \frac{\partial \mathbf{R} }{\partial \psi } =vec ( \mathbf{Z} + \mathbf{Z}^{H} ) \end{aligned}$$

(55)

where

$$\begin{aligned} \mathbf{Z}=(\mathbf{R}^{-1})^{1/2} \mathbf{A} \mathbf{R}_{\tilde{s}} (\mathbf{A}^{^{\prime }})^{H} (\mathbf{R}^{-1})^{1/2}. \end{aligned}$$

(56)

In a similar way to (13) of [14], one obtains the Fisher information scalar

$$\begin{aligned} F_{u} = K \mathbf{g}^{H} \mathbf{P}_{\varvec{\varDelta }}^{\perp } \mathbf{g} \end{aligned}$$

(57)

where $\varvec{\varDelta }=[ \frac{\partial vec(\mathbf{R}) }{\partial \varvec{\rho }}, \frac{\partial vec(\mathbf{R}) }{\partial \sigma ^{2}} ]$ and $\varvec{\rho }$ is a column vector containing free (real) parameters in the conjugate symmetric matrix $\mathbf{R}_{\tilde{s}}$. Note that $\varvec{\varDelta }$ has the same structure as in [14]. Thus (29) of [14] (without subscripts $k,p$ for $\mathbf{Z}$) is also valid for CFO estimation and

$$\begin{aligned} F_{u} =2 K \mathfrak R \left\{ tr \left( \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp } \mathbf{Z}^{H} \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp } \mathbf{Z}^{H} \right) + tr \left( \mathbf{Z} \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp } \mathbf{Z}^{H} \right) \right\} . \end{aligned}$$

(58)

From $\mathbf{Z}^{H} \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp }=\mathbf{0}$,

$$\begin{aligned} F_{u} = 2 K \mathfrak R \left\{ tr \left( \mathbf{Z} \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp } \mathbf{Z}^{H} \right) \right\} . \end{aligned}$$

(59)

Using (23), one can write $\mathbf{Z}^{H}=(\mathbf{R}^{-1})^{1/2} \mathbf{A}^{^{\prime }} \mathbf{R}_{\tilde{s}} \mathbf{A}^{H} (\mathbf{R}^{-1})^{1/2}$. Substituting $\mathbf{Z}$ and thus

$$\begin{aligned} F_{u} = 2 K \mathfrak R \left\{ tr \left( (\mathbf{A}^{^{\prime }})^{H} (\mathbf{R}^{-1})^{1/2} \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp } (\mathbf{R}^{-1})^{1/2} \mathbf{A}^{^{\prime }} \cdot \mathbf{R}_{\tilde{s}} \mathbf{A} \mathbf{R}^{-1} \mathbf{A} \mathbf{R}_{\tilde{s}} \right) \right\} . \end{aligned}$$

(60)

Using (31) in [14],

$$\begin{aligned} (\mathbf{R}^{-1})^{1/2} \mathbf{P}_{ (\mathbf{R}^{-1})^{1/2} \mathbf{A} }^{\perp } (\mathbf{R}^{-1})^{1/2} =\mathbf{P}_{ \mathbf{A} }^{\perp } \mathbf{R}^{-1}. \end{aligned}$$

(61)

Making use of $\mathbf{P}_\mathbf{A}^{\perp } \mathbf{R}^{-1} = (1/\sigma ^{2}) \mathbf{P}_\mathbf{A}^{\perp }$ (after (31) in [14]) and (9),

$$\begin{aligned} \mathbf{P}_{ \mathbf{A} }^{\perp } \mathbf{R}^{-1}= (1/\sigma ^{2}) \mathbf{P}_{ \mathbf{A} }^{\perp } = (1/\sigma ^{2}) (\mathbf{I}-\mathbf{A} \mathbf{A}^{H}). \end{aligned}$$

(62)

Using (61) and (62) in (60) gives that

$$\begin{aligned} F_{u} = (2K /\sigma ^{2}) \mathfrak R \left\{ tr \left( \mathbf{H} \cdot \mathbf{R}_{\tilde{s}} \mathbf{A} \mathbf{R}^{-1} \mathbf{A} \mathbf{R}_{\tilde{s}} \right) \right\} . \end{aligned}$$

(63)

Let the eigen-decomposition of $\mathbf R$ as $\mathbf{R}=\mathbf{U}_{s} \varvec{\varLambda }_{s} \mathbf{U}_{s}^{H} + \sigma ^{2} \mathbf{U}_{n} \mathbf{U}_{n}^{H}$ where $\varvec{\varLambda }_{s}$ is a diagonal matrix containing the $P$ largest eigenvalues of $\mathbf{R}, \sigma ^{2}$ is the value of the remaining (identical) $N-P$ eigenvalues, $\mathbf{U}_{s}$ and $\mathbf{U}_{n}$ contain the corresponding eigenvectors. The dimensions of $\mathbf{U}_{s}$ and $\mathbf{U}_{n}$ are $N \times P$ and $N \times (N-P)$ respectively. Eigenvectors are chosen such that $[\mathbf{U}_{s},\mathbf{U}_{n}]$ is a unitary matrix. Then $\mathbf{R}^{-1}=\mathbf{U}_{s} \varvec{\varLambda }_{s}^{-1} \mathbf{U}_{s}^{H} + \sigma ^{-2} \mathbf{U}_{n} \mathbf{U}_{n}^{H}$. Note that $\mathbf{A}^{H} \mathbf{U}_{s}$ is a $P \times P$ nonsingular matrix and $\mathbf{A}^{H} \mathbf{U}_{n}=\mathbf{0}$. Thus $\mathbf{A}^{H} \mathbf{R}^{-1} \mathbf{A} = \mathbf{A}^{H} \mathbf{U}_{s} \cdot \varvec{\varLambda }_{s}^{-1} \cdot \mathbf{U}_{s}^{H} \mathbf{A}$ where all the three matrices have a full rank $P$. Now it is easy to see that $\mathbf{R}_{\tilde{s}} \mathbf{A} \mathbf{R}^{-1} \mathbf{A} \mathbf{R}_{\tilde{s}}$ has a full rank as well, and the trace value in (63) is positive according to Lemma 1. Then the symbol $\mathfrak R$ therein can be dropped and the lemma is proved from $CRB_{u}=F_{u}^{-1}$.

Appendix 2: Analysis of MUSIC

1.1 The First-order and Second-order Derivatives

Since $\tilde{\mathbf{W}} \tilde{\mathbf{W}}^{H}=\mathbf{I}_{N} - \mathbf{W} \mathbf{W}^{H}$ and $\mathbf{E}_{\theta } \mathbf{E}_{\theta }^{*}=\mathbf{I}_{N}$, one can rewrite the MUSIC function as

$$\begin{aligned} f_{mu}(\theta ) = \frac{1}{K} \sum _{k=1}^{K} \hat{\mathbf{y}}(k)^{H} (\mathbf{I}_{N} - \mathbf{E}_{\theta } \mathbf{W} \mathbf{W}^{H} \mathbf{E}_{\theta }^{*}) \hat{\mathbf{y}}(k). \end{aligned}$$

(64)

The first-order derivative of the MUSIC function (34) is given by

$$\begin{aligned} f_{mu}^{^{\prime }}(\psi )= - \frac{1}{K} \sum _{k=1}^{K} \left\{ \hat{\mathbf{y}}(k)^{H} \mathbf{E}^{^{\prime }} \mathbf{W} \mathbf{W}^{H} \mathbf{E}^{*} \hat{\mathbf{y}}(k) + \hat{\mathbf{y}}(k)^{H} \mathbf{E} \mathbf{W} \mathbf{W}^{H} \mathbf{E}^{^{\prime }*} \hat{\mathbf{y}}(k) \right\} \end{aligned}$$

(65)

Let

$$\begin{aligned} \varvec{\varGamma } =diag[0,1,2,\cdots ,N-1]. \end{aligned}$$

(66)

One can write

$$\begin{aligned} \mathbf{E}^{^{\prime }}&= j \varvec{\varGamma } \mathbf{E} \end{aligned}$$

(67)

$$\begin{aligned} \mathbf{A}^{^{\prime }}&= \mathbf{E}^{^{\prime }} \mathbf{W}= j \varvec{\varGamma } \mathbf{E} \mathbf{W} = j \varvec{\varGamma } \mathbf{A}. \end{aligned}$$

(68)

Define $\dot{\mathbf{A}} = [ j\varvec{\varGamma } \mathbf{A},\mathbf{A}]$ and

$$\begin{aligned} \mathbf{J}= \left[\begin{array}{cc} \mathbf{0}&\mathbf{I}_{P} \\ \mathbf{I}_{P}&\mathbf{0}\\ \end{array} \right]_{2P \times 2P}. \end{aligned}$$

(69)

The first-order derivative can be written in the following form:

$$\begin{aligned} f_{mu}^{^{\prime }}(\psi )&= - \frac{1}{K} \sum _{k=1}^{K} \hat{\mathbf{y}}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \hat{\mathbf{y}}(k) = - \frac{1}{K} \sum _{k=1}^{K} \left\{ \mathbf{y}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{y}(k) \right. \nonumber \\&\left. +\mathbf{y}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) +\mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{y}(k) +\mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \right\} . \end{aligned}$$

(70)

Note that

$$\begin{aligned} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} = j \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} - j \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } \end{aligned}$$

(71)

and using (9)

$$\begin{aligned} \mathbf{y}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{y}(k)&= \tilde{\mathbf{s}}(k)^{H} \mathbf{A}^{H} ( j \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} - j \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } ) \mathbf{A} \tilde{\mathbf{s}}(k) \nonumber \\&= j\tilde{\mathbf{s}}(k)^{H} ( \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} -\mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} ) \tilde{\mathbf{s}}(k) =0. \end{aligned}$$

(72)

Then one has

$$\begin{aligned} f_{mu}^{^{\prime }}(\psi ) = - \frac{1}{K} \sum _{k=1}^{K} \left\{ \mathbf{y}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) +\mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{y}(k) +\mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \right\} . \end{aligned}$$

(73)

The second-order derivative is given by

$$\begin{aligned} f^{^{\prime \prime }}_{mu}(\psi )&= - \frac{2}{K} \mathfrak R \left\{ \hat{\mathbf{y}}(k)^{H} (\mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} ) \hat{\mathbf{y}}(k) \right\} \nonumber \\&= - 2 \mathfrak R \left\{ tr [ (\mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} ) \left( \sum _{k=1}^{K} \hat{\mathbf{y}}(k) \hat{\mathbf{y}}(k)^{H}/K\right)] \right\} \end{aligned}$$

(74)

and its limiting version by

$$\begin{aligned} f^{^{\prime \prime }}_{mu,\infty }(\psi )&= -2 \mathfrak R \left\{ tr [ (\mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} ) ( \mathbf{A} \mathbf{R}_{\tilde{\mathbf{s}}} \mathbf{A}^{H} ) ] \right\} \nonumber \\&- 2 \sigma ^{2} \mathfrak R \{ tr [ \mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} ]\} \end{aligned}$$

(75)

where $\sum _{k=1}^{K} \hat{\mathbf{y}}(k) \hat{\mathbf{y}}(k)^{H}/K$ is approximated by its asymptotic matrix as given in (19). Note that $\mathbf{A}{^{\prime \prime }}= -\varvec{\varGamma }^{2} \mathbf{A}$ and $\mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} =-\varvec{\varGamma }^{2} \mathbf{A} \mathbf{A}^{H} + \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma }$. Thus

$$\begin{aligned}&tr [ \mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} ] =0\end{aligned}$$

(76)

$$\begin{aligned}&tr [ (\mathbf{A}{^{\prime \prime }} \mathbf{A}^{H}+ \mathbf{A}^{^{\prime }} (\mathbf{A}^{^{\prime }})^{H} ) ( \mathbf{A} \mathbf{R}_{\tilde{\mathbf{s}}} \mathbf{A}^{H} ) ] \nonumber \\&=- tr [ (\mathbf{A}^{H} \varvec{\varGamma }\varvec{\varGamma } \mathbf{A} - \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A}) \mathbf{R}_{\tilde{\mathbf{s}}} ] =- tr [ \mathbf{H} \mathbf{R}_{\tilde{\mathbf{s}}} ] \end{aligned}$$

(77)

and thus

$$\begin{aligned}&f^{^{\prime \prime }}_{mu,\infty }(\psi ) = 2 \mathfrak R \{ tr [ \mathbf{H} \mathbf{R}_{\tilde{\mathbf{s}}} ] \} = 2 tr [ \mathbf{H} \mathbf{R}_{\tilde{\mathbf{s}}} ] \end{aligned}$$

(78)

where the symbol $\mathfrak R$ is dropped because the trace value is real (in fact, positive) according to Lemma 1.

1.2 The Mean and Variance of the First-order Derivative

The expectation of the first two terms of (73) are zero because the noise variables have zero means. The third term is

$$\begin{aligned} E \{ \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \} = \sigma ^{2} tr [ \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} ] = \sigma ^{2} tr [ \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} - \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } ] = 0. \end{aligned}$$

Thus

$$\begin{aligned} E \{ f_{mu}^{^{\prime }}(\psi ) \} =0. \end{aligned}$$

(79)

It is well-known that the third-order moment of Gaussian random variables is zero. Using this property and (11), one obtains

$$\begin{aligned} E (f_{mu}^{^{\prime }}(\psi ) )^{2}&= \frac{2}{K^{2} } \sum _{k=1}^{K} E \left\{ \mathbf{y}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{y}(k) \right\} \nonumber \\&+ \frac{1}{K^{2} } \sum _{k=1}^{K} E \left\{ \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \right\} \nonumber \\&= \frac{2 \sigma ^{2}}{K } tr [ \mathbf{A}^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{A} \cdot \hat{\mathbf{R}}_{\tilde{\mathbf{s}}(k)} ] \end{aligned}$$

(80)

$$\begin{aligned}&+ \frac{1}{K} E \left\{ \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \right\} . \end{aligned}$$

(81)

From (71), $\mathbf{A}^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \!= \!j ( \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \!-\! \mathbf{A}^{H} \varvec{\varGamma }) \!=\! - j \mathbf{A}^{H} \varvec{\varGamma } ( \mathbf{I}_{N} \!-\! \mathbf{A} \mathbf{A}^{H} )$, then $\mathbf{A}^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{A} = \mathbf{A}^{H} \varvec{\varGamma } ( \mathbf{I}_{N} - \mathbf{A} \mathbf{A}^{H} ) \varvec{\varGamma } \mathbf{A}$ and thus the term (80) is equal to $\frac{2 \sigma ^{2}}{K} tr [ \mathbf{H} \hat{\mathbf{R}}_{\tilde{\mathbf{s}}} ]$.

An expression of the term (81) can be developed based on the formula (2.4) in [8], for the product of four matrix-valued Gaussian random variables. In (81), only vector-valued Gaussian random variables are involved and a simplified expression is given in the following lemma.

Lemma B1

Let $\mathbf{z}_{i},i=1,2,3,4$ be column vectors of complex Gaussian random variables, with the properties that $E\{ \mathbf{z}_{i} \}= \mathbf{0}$ for some $i\in [1,4], E\{ \mathbf{z}_{3} \otimes \mathbf{z}_{1} \}= \mathbf{0}$ and $E\{ \mathbf{z}_{4} \otimes \mathbf{z}_{2} \}= \mathbf{0}$. Then

$$\begin{aligned} E \{ \mathbf{z}_{1}^{H} \mathbf{z}_{2} \mathbf{z}_{3}^{H} \mathbf{z}_{4}\} = E \{ \mathbf{z}_{1}^{H} \mathbf{z}_{2} \} E \{ \mathbf{z}_{3}^{H} \mathbf{z}_{4}\} + tr [ E \{ \mathbf{z}_{2} \mathbf{z}_{3}^{H} \} E \{ \mathbf{z}_{4} \mathbf{z}_{1}^{H} \} ]. \end{aligned}$$

(82)

For the term (81), $\mathbf{z}_{1}^{H}=\mathbf{n}(k)^{H} \dot{\mathbf{A}}, \mathbf{z}_{2}=\mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k), \mathbf{z}_{3}^{H}=\mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J}, \mathbf{z}_{4}=\dot{\mathbf{A}}^{H} \mathbf{n}(k)$. It is easy to verify that the four vectors satisfy the three properties required in the above lemma. Thus, using Lemma B1, that term is equal to

$$\begin{aligned}&E \{ \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \} E \{ \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \} \nonumber \\&+ tr [ E \{ \mathbf{J} \dot{\mathbf{A}}^{H} \mathbf{n}(k) \mathbf{n}(k)^{H} \dot{\mathbf{A}} \mathbf{J} \} E \{ \dot{\mathbf{A}}^{H} \mathbf{n}(k) \mathbf{n}(k)^{H} \dot{\mathbf{A}} \} ] \nonumber \\&= \sigma ^{4} ( tr [ \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} ] )^{2} + \sigma ^{4} tr [ \mathbf{J} \dot{\mathbf{A}}^{H} \dot{\mathbf{A}} \mathbf{J} \cdot \dot{\mathbf{A}}^{H} \dot{\mathbf{A}} ] \nonumber \\&= \sigma ^{4} ( tr [ \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} ] )^{2} + \sigma ^{4} tr [ \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \cdot \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} ]. \end{aligned}$$

(83)

$tr [ \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} ]=0$ was shown in the proof of (79). Using (71) again,

$$\begin{aligned}&tr [ \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} \cdot \dot{\mathbf{A}} \mathbf{J} \dot{\mathbf{A}}^{H} ]\nonumber \\&= - tr [ \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} - \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } - \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} + \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } ] \nonumber \\&= 2 tr [ \mathbf{A}^{H} \varvec{\varGamma }\varvec{\varGamma } \mathbf{A} - \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A} \mathbf{A}^{H} \varvec{\varGamma } \mathbf{A}] = 2 tr [ \mathbf{H} ]. \end{aligned}$$

(84)

Hence the term (81) is equal to $\frac{2 \sigma ^{4}}{K} tr [ \mathbf{H} ]$. Therefore

$$\begin{aligned} E (f_{mu}^{^{\prime }}(\psi ) )^{2}= \frac{2 \sigma ^{2}}{K } tr [ \mathbf{H} ( \hat{\mathbf{R}}_{\tilde{\mathbf{s}}(k)} + \sigma ^{2} \mathbf{I}_{P} ) ]. \end{aligned}$$

(85)